# Isosceles right triangle problem

• Feb 3rd 2010, 12:00 PM
icu812
Isosceles right triangle problem
Isosceles right triangle problem (solids with know cross-sections)

I have a isosceles right triangle with a known base (square root of (4-x^2) how can I find the height? Do I need to use a law of sin/cos or is there an easier way?
• Feb 3rd 2010, 01:00 PM
Calculus26
The height is the length of one of the equal sides the area is 1/2s^2

since it is a rt triangle s^2 +s^2 = h^2

s= h/sqrt(2)

in your eg h = 4-x^2
• Feb 3rd 2010, 01:02 PM
Soroban
Hello, icu812!

Did you make a sketch?

Quote:

I have a isosceles right triangle with a known base:\ . $b \,=\,\sqrt{4-x^2}$

How can I find the height?

Code:

                  *                 * | *               *  |  *             *    |b/2  *           *      |      *         * 45°    |    45° *       *  *  *  *  *  *  *  *  *       : - b/2 - - : - - b/2 - :       : - - - - - b - - - - - :

Got it?

• Feb 3rd 2010, 01:12 PM
icu812
Yes thanks a lot. It has been probaly 15 years since I had geometry. We just did solids of revolution, and started solids with known cross-section today. (Itwasntme)